我的数据库中有两个表,它与其他表有一对多连接,如下所示
在kvves_units表中,我会得到' name
'来自GET
方法
现在我希望根据kvves_units
kvves_members
和kvves_units
的所有价值
我正在使用类似的代码
$kvvesDetails = $conn->prepare( "SELECT u.id, u.name, u.phone, u.email, u.address, m.name, m.designantion, m.phone, m.email, m.imageFROM kvves_units AS u JOIN kvves_members AS m ON m.unit_id = u.id WHERE `name` = $committee");
答案 0 :(得分:1)
这是标准联接:
$kvvesDetails = $conn->prepare( "SELECT u.id, u.name, u.phone, u.email, u.address, m.name, m.designantion, m.phone, m.email, m.image FROM kvves_units AS u JOIN kvves_members AS m ON m.unit_id = u.id WHERE name = '$committee'"
答案 1 :(得分:1)
使用此SQL
select kvves_units.*,kvves_members.* from kvves_units a join kvves_members b where a.name = b.name and a.name = '".$_GET['name']."'
答案 2 :(得分:0)
尝试:
$name = $_GET['name'];
$sql = "Select *from kvves_units as u INNER JOIN kvves_members as m where u.id = m.unit_id and u.name = '".$name."'";
您将获得解决方案。